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1=1^2 ;125=5^3 ;-125=-5^3; 27=27^1; -27= -27^1
2: 5^2 25^1
Bài 6:
a: \(2^{27}=8^9\)
\(3^{18}=9^9\)
b: Vì \(8^9< 9^9\)
nên \(2^{27}< 3^{18}\)
a) 8 = 23
425 = 25.35.75
16 = 24
b) (0,09)3 = (3/10)6
(3/10)8 = (3/10)8
0,027 = (3/10)3
`@` `\text {Ans}`
`\downarrow`
`a)`
`8 = 2^3`
`32^5` chứ ạ?
`32^5 = (2^5)^5 = 2^10`
`16 = 2^4`
`b)`
`(0,09)^3 = (0,3^2)^3 = 0,3^6` hay `(3/10)^6`
`(3/10)^8 = (3/10)^8`
`(0,027) = (0,3)^3` hay `(3/10)^3`
`@` `\text {Kaizuu lv uuu}`
Ta có: +) \({({2^2})^3} = {2^2}{.2^2}{.2^2} = {2^{2 + 2 + 2}} = {2^6}\)
+) \({\left[ {{{( - 3)}^2}} \right]^2} = {( - 3)^2}.{( - 3)^2} = {( - 3)^{2 + 2}} = {( - 3)^4}\)
\(\left(\dfrac{1}{27}\right)^5\) = \(\left(\dfrac{1}{3^3}\right)^5\) = \(\left(\dfrac{1}{3}\right)^{15}\)
\(\left(\dfrac{1}{27}\right)^5=\left[\left(\dfrac{1}{3}\right)^3\right]^5=\dfrac{1}{3}^{3.5}=\dfrac{1}{3}^{15}\)
Ta có:
\(\begin{array}{l}{\left( {\frac{1}{9}} \right)^5} = {[{\left( {\frac{1}{3}} \right)^2}]^5} = {(\frac{1}{3})^{2.5}} = {(\frac{1}{3})^{10}};\\{\left( {\frac{1}{{27}}} \right)^7} = {[{(\frac{1}{3})^3}]^7} = {(\frac{1}{3})^{3.7}} = {(\frac{1}{3})^{21}}\end{array}\)
Ta có:
\(\begin{array}{l}{\left( {\frac{1}{4}} \right)^8} = {[{\left( {\frac{1}{2}} \right)^2}]^8} = {(\frac{1}{2})^{2.8}} = {(\frac{1}{2})^{16}};\\{\left( {\frac{1}{8}} \right)^3} = {[{(\frac{1}{2})^3}]^3} = {(\frac{1}{2})^{3.3}} = {(\frac{1}{2})^9}\end{array}\)
\(-\dfrac{8}{27}=\left(-\dfrac{2}{3}\right)^3\)
\(\dfrac{81}{625}=\left(\dfrac{3}{5}\right)^4\)
\(\dfrac{25}{49}=\left(\dfrac{5}{7}\right)^2\)
\(\frac{25.5^3.1}{625.5^2}=\frac{5^2.5^3}{5^4.5^2}=\frac{1}{5}\)
các câu sau mk k hiểu bn viết j
Nếu đề bài là: (\(\dfrac{2}{3}\))5 x \(\dfrac{1}{2^5}\) thì làm như này em nhé.
(\(\dfrac{2}{3}\))5 x \(\dfrac{1}{2^5}\) = (\(\dfrac{2}{3}\) x \(\dfrac{1}{2}\))5 = (\(\dfrac{1}{3}\))5